Kinetic Theory Of Gases Question 56
Question: At $ 10{}^\circ C $ the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At $ 110{}^\circ C $ this ratio is:
Options:
A) x
B) $ \frac{383}{283}x $
C) $ \frac{10}{110}x $
D) $ \frac{283}{383}x $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let the mass of the gas be m. At a fixed temperature and pressure, volume is fixed. Density of the gas, $ \rho =\frac{m}{V} $ Now $ \frac{\rho }{P}=\frac{m}{PV}=\frac{m}{nRT} $
$ \Rightarrow \frac{m}{nRT}=x( \text{By question} ) $
$ \Rightarrow xT=\text{constant}\Rightarrow x _{1}T _{1}=x _{2}T _{2} $
$ \Rightarrow x _{2}\Rightarrow \frac{x _{1}T _{1}}{T _{2}}=\frac{283}{383}\times [ \
Therefore T _{1}=283K T _{2}=383K ] $
BETA
